Bitwise AND
 

This is giving me a headache...can someone please tell me why these statements ARE NOT equivalent?

Statement 1.)

if(running_result && shell[8 * (x - 1) + (y - 1)].input_val)
running_result = true;
else
running_result = false;


Statement 2.)

running_result &= shell[8 * (x - 1) + (y - 1)].input_val;

-------------------------
The first statement works; the second does NOT. The second always evaluates to false. Why? Both of these vars are bools.

Re: Bitwise AND
 

Maybe the problem is elsewhere.

#include <iostream>

  

  int main()

  {

     bool a, b;

  

     //Statement 1.)

     a = false; b = false; std::cout << "a = " << a << ", b = " << b;

     if ( a && b )

            a = true;

     else

            a = false;

     std::cout << ": a = " << a << std::endl;

  

     a = false; b = true; std::cout << "a = " << a << ", b = " << b;

     if ( a && b )

            a = true;

     else

            a = false;

     std::cout << ": a = " << a << std::endl;

  

     a = true; b = false; std::cout << "a = " << a << ", b = " << b;

     if ( a && b )

            a = true;

     else

            a = false;

     std::cout << ": a = " << a << std::endl;

  

     a = true; b = true; std::cout << "a = " << a << ", b = " << b;

     if ( a && b )

            a = true;

     else

            a = false;

     std::cout << ": a = " << a << std::endl;

     std::cout << std::endl;

  

     //Statement 2.)

     a = false; b = false; std::cout << "a = " << a << ", b = " << b;

     a &= b;

     std::cout << ": a = " << a << std::endl;

  

     a = false; b = true; std::cout << "a = " << a << ", b = " << b;

     a &= b;

     std::cout << ": a = " << a << std::endl;

  

     a = true; b = false; std::cout << "a = " << a << ", b = " << b;

     a &= b;

     std::cout << ": a = " << a << std::endl;

  

     a = true; b = true; std::cout << "a = " << a << ", b = " << b;

     a &= b;

     std::cout << ": a = " << a << std::endl;

     std::cout << std::endl;

  

     return 0;

  }

  

  /* my output

  a = 0, b = 0: a = 0

  a = 0, b = 1: a = 0

  a = 1, b = 0: a = 0

  a = 1, b = 1: a = 1

  

  a = 0, b = 0: a = 0

  a = 0, b = 1: a = 0

  a = 1, b = 0: a = 0

  a = 1, b = 1: a = 1

  */

Re: Bitwise AND
 

the first statements uses logical and (&&) the second uses bitwise and (&), so you might try using '&&=' in the second statement.

The difference might be subtle; say the input_val was '2'. In the first statement the running result would be true and in the second statement it would be 0 because TRUE (1) & 2 == 0, but (1 && 2) is true.

Re: Bitwise AND
 

There is no such critter: '&&='.

Re: Bitwise AND
 

Hee hee - well there SHOULD be! In any case, & and && are not the same.

Re: Bitwise AND
 

But for bools, the results should be the same, shoudn't they?

Re: Bitwise AND
 

I am so excited because I finished my program.

Chainsaw is right, any nonzero value is true in a bool, so a bitwise AND wouldn't necessarily evaluate to true... and I shouldn't have tried using a bitwise op for boolean algebra.

(Incidentally I settled on):

running_result = running_result && shell[8 * (x - 1) + (y - 1)].input_val;

Not too stylish, but it served its purpose.